How can I explicitly express the Ising Hamiltonian in matrix form?
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
|
show 14 more comments
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
1
Should be $H_{rs}$
– InertialObserver
5 hours ago
1
What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago
1
Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago
1
I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago
2
Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago
|
show 14 more comments
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
I am reading this book about numerical methods in physics. It has the following question:
Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$
Write a program that computes the $2^N times 2^N$ matrix for
different $N$
From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.
My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?
condensed-matter quantum-spin computational-physics ising-model spin-chains
condensed-matter quantum-spin computational-physics ising-model spin-chains
edited 4 hours ago
Norbert Schuch
8,42122438
8,42122438
asked 5 hours ago
Luqman Saleem
249111
249111
1
Should be $H_{rs}$
– InertialObserver
5 hours ago
1
What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago
1
Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago
1
I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago
2
Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago
|
show 14 more comments
1
Should be $H_{rs}$
– InertialObserver
5 hours ago
1
What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago
1
Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago
1
I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago
2
Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago
1
1
Should be $H_{rs}$
– InertialObserver
5 hours ago
Should be $H_{rs}$
– InertialObserver
5 hours ago
1
1
What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago
What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago
1
1
Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago
Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago
1
1
I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago
I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago
2
2
Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago
Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago
|
show 14 more comments
2 Answers
2
active
oldest
votes
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$
as
$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
|
show 3 more comments
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.
2
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
active
oldest
votes
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$
as
$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
|
show 3 more comments
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$
as
$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
|
show 3 more comments
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$
as
$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.
For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write
$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$
as
$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$
where it is understood that (to prevent clutter)
$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$
and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)
We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices
$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$
We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.
edited 3 hours ago
answered 4 hours ago
InertialObserver
1,300517
1,300517
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
|
show 3 more comments
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
Will do, gonna do that now
– InertialObserver
3 hours ago
1
1
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago
|
show 3 more comments
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.
2
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
add a comment |
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.
2
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
add a comment |
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.
$textbf{Quick hint :}$
Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):
$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices
Note also that Kronecker product is associative.
$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.
edited 3 hours ago
answered 4 hours ago
Sunyam
6021311
6021311
2
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
add a comment |
2
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
2
2
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
That was very helpful. Thank you.
– Luqman Saleem
4 hours ago
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1
Should be $H_{rs}$
– InertialObserver
5 hours ago
1
What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago
1
Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago
1
I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago
2
Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago